A pinch of Physics is required!

You have a piece of aluminium whose mass is 800 g 800\text{ g} , and suppose it is heated up to 100 0 C . 1000^\circ\text{C}.

Now, given that you are cooling it down to 15 0 C 150^\circ\text{C} , calculate the amount of heat (in Joules) given out during the process of cooling.

Details and assumptions:

  • The specific heat capacity of aluminium is C Aluminum = 900 J / kg K . \text{C}_\text{Aluminum}=900 \text{ J}/\text{kg K}.
  • Ignore the heat energy lost to the atmosphere.

Image credit : hardy solutions


The answer is 612000.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Sravanth C.
May 19, 2015

The following values are given,

m a s s = 800 g = 0.8 k g I n i t i a l t e m p e r a t u r e = T 1 = 1000 ° C = 1273 ° K F i n a l t e m p e r a t u r e = 150 ° C = 423 ° K C a l u m i n i u m = 900 J / k g K mass=800g=0.8kg \\ Initial \quad temperature=T_{1}=1000°C=1273°K \\ Final \quad temperature =150°C=423°K \\ C_{aluminium}=900J/kgK

We know that heat energy Q Q is given by the following expression,

Q = C m Δ T = 900 × 0.8 × ( 1273 423 ) = 720 × 850 = 612000 J Q=Cm\Delta T \\ =900×0.8×(1273-423) \\ =720×850 \\ =612000J

I think you meant Δ T \Delta T .BTW

C h e e r s ! \huge Cheers!

Nihar Mahajan - 6 years ago

Log in to reply

Thanks! Edited.

Sravanth C. - 6 years ago
Nihar Mahajan
May 19, 2015

My solution is almost similar to that of Sravanth but I like celsius more than kelvin.

C a l u m i n i u m = 900 J / k g K = 900 J / k g C C_{aluminium}=900J/kg K =900J/kg C \ because a unit change in celsius is equal to unit change in kelvin.So the formula for release of heat is given by :

Δ Q = m . C a l u m i n i u m Δ T = ( 0.8 ) ( 900 ) ( 1000 150 ) = 612000 J \Delta Q = m.C_{aluminium} \Delta T = (0.8)(900)(1000-150)=612000 J

Oh yeah! The change in temperatures in both the scales is the same.

C H E E R S ! ! ! \huge CHEERS!!!

Sravanth C. - 6 years ago

Log in to reply

¨ ¨ ¨ \ddot\smile\ddot\smile\ddot\smile

Nihar Mahajan - 6 years ago

Have some room for me back there guys? xD

Arian Tashakkor - 6 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...